A 5 kg slender rod of length 10 m is suspended from the pin at A and is at rest.

Question

                    A 5 kg slender rod of length 10 m is suspended from the pin at A and is at rest. A 0.5 kg Ball B impacts the slender rod at its Center of Mass with a velocity                     of 20 m/sec. The coefficient of restitution between the rod and ball is 0.5. Determine the angular velocity of the rod and the velocity of Ball B immediately                     after impact                 

                    
                

                    
                                     

A 5 kg slender rod of length 10 m is suspended from the pin at A and is at rest. A 0.5 kg Ball B impacts the slender rod at its Center of Mass with a velocity of 20 m/sec. The coefficient of restitution between the rod and ball is 0.5. Determine the angular velocity of the rod and the velocity of Ball B immediately after impact

Explanation / Answer

M = 5 kg

m = 0.5 kg

L = 10 m

v1 = 20 m/s

e = 0.5

r = L/2 = 10/2 = 5 m

w = angular velocity of bar after impact

v_bar = rw = 5w

e = (v_bar - v2) / (v1 - 0)..............1

Also, momentum conservation gives m*v1 = M*v_bar + m*v2

or m(v1 - v2)/M = v_bar...........2

Puting eqn 2 in 1, e = [m(v1 - v2)/M - v2] / v1

e*v1 + v2 = m*v1/M - m*v2/M

v1*(m/M - e) = v2*(1 + m/M)

v2 = v1*(m/M - e) / (1 + m/M)

v2 = 20*(0.5 / 5 - 0.5) / (1 + 0.5/5)

v2 = -7.27 m/s

Thus, substituting it back in eqn 1,

0.5 = (v_bar - (-7.27)) / 20

v_bar = 2.727 m/s

w = v_bar / r

w = 2.727 / 5

w = 0.545 rad/s

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